YES

We show the termination of the TRS R:

  f(x,f(a(),y)) -> f(f(y,f(f(a(),a()),a())),x)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(x,f(a(),y)) -> f#(f(y,f(f(a(),a()),a())),x)
p2: f#(x,f(a(),y)) -> f#(y,f(f(a(),a()),a()))
p3: f#(x,f(a(),y)) -> f#(f(a(),a()),a())
p4: f#(x,f(a(),y)) -> f#(a(),a())

and R consists of:

r1: f(x,f(a(),y)) -> f(f(y,f(f(a(),a()),a())),x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(x,f(a(),y)) -> f#(f(y,f(f(a(),a()),a())),x)

and R consists of:

r1: f(x,f(a(),y)) -> f(f(y,f(f(a(),a()),a())),x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{x1 + 25, x2 + 24}
        f_A(x1,x2) = max{1, x1 - 12, x2 - 8}
        a_A = 15
  
    precedence:
    
      f# > a > f
    
    partial status:
  
      pi(f#) = []
      pi(f) = []
      pi(a) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.